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Anything fermionic.

When you Wick rotate, the time direction becomes compact; a circle of length $\beta$, where $\beta$ is the inverse temperature. As such, you have to choose boundary conditions for your fields when they go around the time circle. For bosonic systems the only natural thing is to choose periodic boundary conditions $\phi(\tau+\beta)=\phi(\tau)$.

However with fermions you can either have periodic (P), or anti-periodic (AP) boundary conditions: \begin{align} \text{P:}\qquad &\psi(\tau+\beta)=\psi(\tau) \\ \text{AP:}\qquad &\psi(\tau+\beta)=-\psi(\tau). \end{align} This fact (along with choices of boundary condition around other compact directions of your base space) is sometimes referred to as a choice of a spin structure. Anyways, you then go on to compute the Euclidean path integral with your choice of spin structure: $$Z_\text{PI}(\text{BC}) := \int_{\text{BC}} \mathrm{D}\psi \ \mathrm{e}^{-S[\psi]}.$$ You can show very explicitly (take e.g. just a $(0+1)$-dimensional free fermion and compute its path integral) that if you choose anti-periodic boundary conditions it gives you exactly the statistical partition function: $$Z_\text{PI}(\text{AP}) = \mathrm{tr}\,\mathrm{e}^{-\beta H}.$$ However, if you choose periodic boundary conditions, it gives you a quantity that cannot be seen as a statistical partition function; the Witten indexthe Witten index: $$Z_\text{PI}(\text{P}) = \mathrm{tr}\,(-1)^F\mathrm{e}^{-\beta H},$$ where $F$ is the fermion number operator.

You might object and say, well, why would I then choose periodic boundary conditions? Sometimes, the symmetries of the system enforce on you the boundary conditions. For example, to keep on with the working example of a free $(0+1)$-dimensional fermion, let me add to it a single, bosonic harmonic oscillator. Now you have an action of the form $$S[q,\psi] = \int_{0}^\beta \mathrm{d}\tau\left(\frac12 \dot{q}^2 + \frac12 \omega^2 q^2 + \bar\psi\dot\psi + \omega\bar\psi\psi\right).$$ You can show that this theory is in fact supersymmetric, if and only if, the fermion, $\psi$ has periodic boundary conditions. So if you wanted to cook up a supersymmetric theory like so, you would have to end up with periodic boundary conditions and a Euclidean path integral that does not compute a statistical partition function.

Anything fermionic.

When you Wick rotate, the time direction becomes compact; a circle of length $\beta$, where $\beta$ is the inverse temperature. As such, you have to choose boundary conditions for your fields when they go around the time circle. For bosonic systems the only natural thing is to choose periodic boundary conditions $\phi(\tau+\beta)=\phi(\tau)$.

However with fermions you can either have periodic (P), or anti-periodic (AP) boundary conditions: \begin{align} \text{P:}\qquad &\psi(\tau+\beta)=\psi(\tau) \\ \text{AP:}\qquad &\psi(\tau+\beta)=-\psi(\tau). \end{align} This fact (along with choices of boundary condition around other compact directions of your base space) is sometimes referred to as a choice of a spin structure. Anyways, you then go on to compute the Euclidean path integral with your choice of spin structure: $$Z_\text{PI}(\text{BC}) := \int_{\text{BC}} \mathrm{D}\psi \ \mathrm{e}^{-S[\psi]}.$$ You can show very explicitly (take e.g. just a $(0+1)$-dimensional free fermion and compute its path integral) that if you choose anti-periodic boundary conditions it gives you exactly the statistical partition function: $$Z_\text{PI}(\text{AP}) = \mathrm{tr}\,\mathrm{e}^{-\beta H}.$$ However, if you choose periodic boundary conditions, it gives you a quantity that cannot be seen as a statistical partition function; the Witten index: $$Z_\text{PI}(\text{P}) = \mathrm{tr}\,(-1)^F\mathrm{e}^{-\beta H},$$ where $F$ is the fermion number operator.

You might object and say, well, why would I then choose periodic boundary conditions? Sometimes, the symmetries of the system enforce on you the boundary conditions. For example, to keep on with the working example of a free $(0+1)$-dimensional fermion, let me add to it a single, bosonic harmonic oscillator. Now you have an action of the form $$S[q,\psi] = \int_{0}^\beta \mathrm{d}\tau\left(\frac12 \dot{q}^2 + \frac12 \omega^2 q^2 + \bar\psi\dot\psi + \omega\bar\psi\psi\right).$$ You can show that this theory is in fact supersymmetric, if and only if, the fermion, $\psi$ has periodic boundary conditions. So if you wanted to cook up a supersymmetric theory like so, you would have to end up with periodic boundary conditions and a Euclidean path integral that does not compute a statistical partition function.

Anything fermionic.

When you Wick rotate, the time direction becomes compact; a circle of length $\beta$, where $\beta$ is the inverse temperature. As such, you have to choose boundary conditions for your fields when they go around the time circle. For bosonic systems the only natural thing is to choose periodic boundary conditions $\phi(\tau+\beta)=\phi(\tau)$.

However with fermions you can either have periodic (P), or anti-periodic (AP) boundary conditions: \begin{align} \text{P:}\qquad &\psi(\tau+\beta)=\psi(\tau) \\ \text{AP:}\qquad &\psi(\tau+\beta)=-\psi(\tau). \end{align} This fact (along with choices of boundary condition around other compact directions of your base space) is sometimes referred to as a choice of a spin structure. Anyways, you then go on to compute the Euclidean path integral with your choice of spin structure: $$Z_\text{PI}(\text{BC}) := \int_{\text{BC}} \mathrm{D}\psi \ \mathrm{e}^{-S[\psi]}.$$ You can show very explicitly (take e.g. just a $(0+1)$-dimensional free fermion and compute its path integral) that if you choose anti-periodic boundary conditions it gives you exactly the statistical partition function: $$Z_\text{PI}(\text{AP}) = \mathrm{tr}\,\mathrm{e}^{-\beta H}.$$ However, if you choose periodic boundary conditions, it gives you a quantity that cannot be seen as a statistical partition function; the Witten index: $$Z_\text{PI}(\text{P}) = \mathrm{tr}\,(-1)^F\mathrm{e}^{-\beta H},$$ where $F$ is the fermion number operator.

You might object and say, well, why would I then choose periodic boundary conditions? Sometimes, the symmetries of the system enforce on you the boundary conditions. For example, to keep on with the working example of a free $(0+1)$-dimensional fermion, let me add to it a single, bosonic harmonic oscillator. Now you have an action of the form $$S[q,\psi] = \int_{0}^\beta \mathrm{d}\tau\left(\frac12 \dot{q}^2 + \frac12 \omega^2 q^2 + \bar\psi\dot\psi + \omega\bar\psi\psi\right).$$ You can show that this theory is in fact supersymmetric, if and only if, the fermion, $\psi$ has periodic boundary conditions. So if you wanted to cook up a supersymmetric theory like so, you would have to end up with periodic boundary conditions and a Euclidean path integral that does not compute a statistical partition function.

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ɪdɪət strəʊlə
ɪdɪət strəʊlə
  • 4.8k
  • 1
  • 12
  • 33

Anything fermionic.

When you Wick rotate, the time direction becomes compact; a circle of length $\beta$, where $\beta$ is the inverse temperature. As such, you have to choose boundary conditions for your fields when they go around the time circle. For bosonic systems the only natural thing is to choose periodic boundary conditions $\phi(\tau+\beta)=\phi(\tau)$.

However with fermions you can either have periodic (P), or anti-periodic (AP) boundary conditions: \begin{align} \text{P:}\qquad &\psi(\tau+\beta)=\psi(\tau) \\ \text{AP:}\qquad &\psi(\tau+\beta)=-\psi(\tau). \end{align} This fact (along with choices of boundary condition around other compact directions of your base space) is sometimes referred to as a choice of a spin structure. Anyways, you then go on to compute the Euclidean path integral with your choice of spin structure: $$Z_\text{PI}(\text{BC}) := \int_{\text{BC}} \mathrm{D}\psi \ \mathrm{e}^{-S[\psi]}.$$ You can show very explicitly (take e.g. just a $(0+1)$-dimensional free fermion and compute its path integral) that if you choose anti-periodic boundary conditions it gives you exactly the statistical partition function: $$Z_\text{PI}(\text{AP}) = \mathrm{tr}\,\mathrm{e}^{-\beta H}.$$ However, if you choose periodic boundary conditions, it gives you a quantity that cannot be seen as a statistical partition function; the Witten index: $$Z_\text{PI}(\text{P}) = \mathrm{tr}\,(-1)^F\mathrm{e}^{-\beta H},$$ where $F$ is the fermion number operator.

You might object and say, well, why would I then choose periodic boundary conditions? Sometimes, the symmetries of the system enforce on you the boundary conditions. For example, to keep on with the working example of a free $(0+1)$-dimensional fermion, let me add to it a single, bosonic harmonic oscillator. Now you have an action of the form $$S[q,\psi] = \int_{0}^\beta \mathrm{d}\tau\left(\frac12 \dot{q}^2 + \frac12 \omega^2 q^2 + \bar\psi\dot\psi + \omega\bar\psi\psi\right).$$ You can show that this theory is in fact supersymmetric, if and only if, the fermion, $\psi$ has periodic boundary conditions. So if you wanted to cook up a supersymmetric theory like so, you would have to end up with periodic boundary conditions and a Euclidean path integral that does not compute a statistical partition function.